If \({\rm{A}} = {\rm{}}\frac{1}{{1{\rm{\;}} \times {\rm{\;}}2}}{\rm{}} + {\rm{}}\frac{1}{{1{\rm{\;}} \times {\rm{\;}}4}}{\rm{}} + {\rm{}}\frac{1}{{2{\rm{\;}} \times {\rm{\;}}3}}{\rm{}} + {\rm{}}\frac{1}{{4{\rm{\;}} \times {\rm{\;}}7}}{\rm{}} + {\rm{}}\frac{1}{{3{\rm{\;}} \times {\rm{\;}}4}}{\rm{}} + {\rm{}}\frac{1}{{7{\rm{\;}} \times {\rm{\;}}10}} - - -\) upto 20 terms, then what is the value of A?
420/341
The 20 terms split into two telescoping series of 10 terms each.
Series 1: \(\sum_{k=1}^{10}\frac{1}{k(k+1)}=\sum\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{11}=\frac{10}{11}\)
Series 2: \(\sum_{k=1}^{10}\frac{1}{(3k-2)(3k+1)}=\frac{1}{3}\left(1-\frac{1}{31}\right)=\frac{10}{31}\)
\[A=\frac{10}{11}+\frac{10}{31}=\frac{310+110}{341}=\frac{420}{341}\]
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Column I
Column II
a.
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i.
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b.
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ii.
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d.
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